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3 years ago

What is the purpose of the wire coil in an electromagnet?

Physics

1 answer:

ziro4ka [17]3 years ago

3 0

It is to conduct electricity in the magnet so it has an electric field.

Please BRAINLIEST!

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A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

$v=\dfrac{kh}{l}$

Put the value into the formula

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0

4 years ago

State an advantage of using such hydraulic jack to lift a load

expeople1 [14]

Answer:

Explanation:

You are going to lift and press down on the 200 N many times and move only a short distance. The reward is that slowly but surely you will lift a very heavy load -- one that cannot be managed any other way but by the hydraulic jack.

5 0

3 years ago

What is the cause of atmospheric pressure.

marusya05 [52]

Explanation:

hope it's help you ok have a good day

6 0

3 years ago

A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground

Oksi-84 [34.3K]

31.3m/s

Explanation:

Given parameters:

Mass of rock = 40kg

Height of cliff = 50m

Unknown:

Speed of rock when it hits ground = ?

Solution:

We are going to use the appropriate motion equation to solve this problem

The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.

Using;

V² = U² + 2gH

V = unknown velocity

U = initial velocity = O

g = acceleration due to gravity = 9.8m/s²

H = height of fall

since the initial velocity of the bodyg is 0

V² = 2gH

V= √2gH = √2 x 9.8 x 50 = 31.3m/s

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

7 0

4 years ago

A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 521.59 Hz

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.

8 0

3 years ago